Question

Nonexercise activity thermogenesis (NEAT) is energy burned by fidgeting, maintenance of posture, spontaneous muscle contraction, and other activities of daily living. In a study, 16 subjects increased their NEAT by 327 calories per day, on average, in response to additional food intake of 1000 calories a day. The standard deviation was 256. (a) Test the null hypothesis that there was no change in NEAT versus the two-sided alternative. State the appropriate Ho. HO:20 O Ho:> 0 AIA OOOOO OHO: NSO State the appropriate Hz. Она: ц> 0 O Haru = 0 O Hallo OOOOO Carry out the test. (Round your answer for t to three decimal places.) Give the degrees of freedom. Give the P-value. (Round your answer to four decimal places.)

Answer 1

(a) Appropriate null hypothesis is

$H_{0}:\mu =0$

Appropriate alternative hypothesis is

  $H_{1}:\mu \neq 0$

  $t= \frac{\overline{x}-\mu }{s/\sqrt{n-1}}$

Here $\overline{x}=327, s=256, n=16$

Thus t = 4.947

Degrees of freedom = n-1 = 15

p-value = 0.0002

Here p-value is less than 0.05. Hence we reject the hull hypothesis.

i.e; we have sufficient evidence to conclude that there was a change in NEAT

95 %  confidence interval for the change in NEAT

$\overline{x}-t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n-1}},\overline{x}+t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n-1}}$

$t_{\frac{\alpha }{2}}=2.131(From ~t-distribution ~tables)$

Hence 95% confidence interval for NEAT is

(186.1, 467.9)